Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $p = \dfrac{r^2 - 2r - 3}{-4r^2 - 4r} \times \dfrac{-5r + 40}{-10r + 30} $
Solution: First factor the quadratic. $p = \dfrac{(r - 3)(r + 1)}{-4r^2 - 4r} \times \dfrac{-5r + 40}{-10r + 30} $ Then factor out any other terms. $p = \dfrac{(r - 3)(r + 1)}{-4r(r + 1)} \times \dfrac{-5(r - 8)}{-10(r - 3)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (r - 3)(r + 1) \times -5(r - 8) } { -4r(r + 1) \times -10(r - 3) } $ $p = \dfrac{ -5(r - 3)(r + 1)(r - 8)}{ 40r(r + 1)(r - 3)} $ Notice that $(r + 1)$ and $(r - 3)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -5\cancel{(r - 3)}(r + 1)(r - 8)}{ 40r(r + 1)\cancel{(r - 3)}} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $p = \dfrac{ -5\cancel{(r - 3)}\cancel{(r + 1)}(r - 8)}{ 40r\cancel{(r + 1)}\cancel{(r - 3)}} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $p = \dfrac{-5(r - 8)}{40r} $ $p = \dfrac{-(r - 8)}{8r} ; \space r \neq 3 ; \space r \neq -1 $